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\(=\dfrac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}-\dfrac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\dfrac{\sqrt{3}+\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}-...+\dfrac{\sqrt{2017}+\sqrt{2018}}{\left(\sqrt{2017}+\sqrt{2018}\right)\left(\sqrt{2017}-\sqrt{2018}\right)}\)
\(=\dfrac{\sqrt{1}+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-...+\dfrac{\sqrt{2017}+\sqrt{2018}}{2017-2018}\)
\(=-\left(\sqrt{1}+\sqrt{2}\right)+\sqrt{2}+\sqrt{3}-\left(\sqrt{3}+\sqrt{4}\right)+...-\left(\sqrt{2017}+\sqrt{2018}\right)\)
\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...-\sqrt{2017}-\sqrt{2018}\)
\(=-\sqrt{1}-\sqrt{2018}\)
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
1)
DKCĐ: a>0,\(a\ne1\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}(n+1-n)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)
\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
=>A là số hữu tỉ (ĐPCM)
Lời giải:
Xét \(1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=\frac{n^2+1}{n^2}+\frac{1}{(n+1)^2}\)
\(=\frac{(n+1)^2-2n}{n^2}+\frac{1}{(n+1)^2}=\left(\frac{n+1}{n}\right)^2+\frac{1}{(n+1)^2}-\frac{2}{n}\)
\(=\left(\frac{n+1}{n}-\frac{1}{n+1}\right)^2=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\)
\(\Rightarrow \sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n}-\frac{1}{n+1}\)
Áp dụng vào bài toán suy ra:
\(A=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=2016+\frac{1}{2}-\frac{1}{2018}=2016,5-\frac{1}{2018}\)
\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)